Problem 25 - Thomas Calculus

Question

Which of the series in Exercises \(11-40\) converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series' convergence or divergence.) $$ \sum_{n=1}^{\infty} \frac{2^{n}}{n+1} $$

Step-by-Step Solution

Answer

The series \( \sum_{n=1}^{\infty} \frac{2^n}{n+1} \) diverges due to the ratio test result being greater than 1.

Step 1: Identify the Type of Series

The given series is \( \sum_{n=1}^{\infty} \frac{2^n}{n+1} \). Notice that this series is not in a form that can directly identify it as a geometric or p-series. However, it has a term involving \(2^n\) which suggests an exponential growth related to terms of a geometric series.

Step 2: Use the Ratio Test

To determine convergence, we apply the ratio test for series. Let's consider \( a_n = \frac{2^n}{n+1} \). This implies that: \( a_{n+1} = \frac{2^{n+1}}{n+2} \). Compute the ratio \( \frac{a_{n+1}}{a_n} \):\[\frac{a_{n+1}}{a_n} = \frac{\frac{2^{n+1}}{n+2}}{\frac{2^n}{n+1}} = \frac{2 \cdot (n+1)}{n+2}\].

Step 3: Evaluate the Limit in the Ratio Test

Compute the limit \( \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{2(n+1)}{n+2} \). Divide the numerator and the denominator by \( n \):\[\lim_{n \to \infty} \frac{2(n+1)}{n+2} = \lim_{n \to \infty} \frac{2 + \frac{2}{n}}{1 + \frac{2}{n}} = 2\].

Step 4: Interpret the Limit Result

Since the result from the ratio test is 2 and the limit obtained is greater than 1, the ratio test shows that the series \( \sum_{n=1}^{\infty} \frac{2^n}{n+1} \) diverges. The ratio test confirms divergence when the limit is greater than 1.

Key Concepts

Ratio Test

The Ratio Test is a powerful tool used to determine whether a series converges or diverges. It is particularly useful when dealing with series that feature exponential growth, such as those with terms like \( a_n = \frac{2^n}{n+1} \).
To apply the Ratio Test, we first define the consecutive terms of a series: \( a_n \) and \( a_{n+1} \). We then compute the ratio \( \frac{a_{n+1}}{a_n} \). Next, we evaluate the limit:

• If \(\lim_{n \to \infty} \frac{a_{n+1}}{a_n} < 1\), the series converges.
• If \(\lim_{n \to \infty} \frac{a_{n+1}}{a_n} > 1\), the series diverges.
• If \(\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = 1\), the test is inconclusive.

In the given step-by-step solution, the limit was found to be 2, which is greater than 1, indicating that the series \( \sum_{n=1}^{\infty} \frac{2^n}{n+1} \) diverges.

Geometric Series

A Geometric Series is a type of series with a constant ratio between consecutive terms. It takes the form \( a + ar + ar^2 + ar^3 + \ldots \), where \( a \) is the first term and \( r \) is the common ratio.

• If \(|r| < 1\), the geometric series converges to \( \frac{a}{1-r} \).
• If \(|r| \geq 1\), the geometric series diverges.

Though the series \( \sum_{n=1}^{\infty} \frac{2^n}{n+1} \) isn't a geometric series due to the variable denominator \( n+1 \), the presence of \( 2^n \) suggests an element of exponential growth similar to geometric series. Understanding this can guide us in selecting suitable convergence tests, such as the Ratio Test.

Exponential Growth

Exponential Growth in the context of series involves terms that grow rapidly with respect to the term index \( n \), like \( 2^n \). Such terms can significantly impact the convergence or divergence of a series due to their fast growth.
In the exercise, the term \( 2^n \) suggests that the series may not simply converge by inspection, prompting the use of tests like the Ratio Test to rigorously establish convergence behavior.
When dealing with exponential growth:

• Terms grow quickly, potentially leading to divergence unless mitigated by other factors like a rapidly increasing denominator.
• Understanding growth patterns helps anticipate the application of specific tests like the Ratio Test.

Being aware of exponential growth reinforces the insight that the series \( \sum_{n=1}^{\infty} \frac{2^n}{n+1} \) diverges.